Answer:
21% of the animals are female
Step-by-step explanation:
In total:
There are 40 + 60 = 100 animals.
Female animals.
40 dogs. Of those, 15% are female. So 0.15*40 = 6.
60 cats. Of those, 25% are female. So 0.25*60 = 15.
21 female animals.
21/100 = 0.21 = 21%
21% of the animals are female
The distance the tip travels is
2π·(40 m) = 80π m
Then its speed is
(80π m)/(10 s) = 8π m/2 ≈ 25.1 m/s
There's not enough given information in the question to calculate an answer.
If we only know that the volume of the box is 96 cm³, then there are an infinite
number of different lengths, widths, and surface areas that it could have.
Answer:
1) The probability that ten students in a class have different birthdays is 0.883.
2) The probability that among ten students in a class, at least two of them share a birthday is 0.002.
Step-by-step explanation:
Given : Assume there are 365 days in a year.
To find : 1) What is the probability that ten students in a class have different birthdays?
2) What is the probability that among ten students in a class, at least two of them share a birthday?
Solution :

Total outcome = 365
1) Probability that ten students in a class have different birthdays is
The first student can have the birthday on any of the 365 days, the second one only 364/365 and so on...

The probability that ten students in a class have different birthdays is 0.883.
2) The probability that among ten students in a class, at least two of them share a birthday
P(2 born on same day) = 1- P( 2 not born on same day)
![\text{P(2 born on same day) }=1-[\frac{365}{365}\times \frac{364}{365}]](https://tex.z-dn.net/?f=%5Ctext%7BP%282%20born%20on%20same%20day%29%20%7D%3D1-%5B%5Cfrac%7B365%7D%7B365%7D%5Ctimes%20%5Cfrac%7B364%7D%7B365%7D%5D)
![\text{P(2 born on same day) }=1-[\frac{364}{365}]](https://tex.z-dn.net/?f=%5Ctext%7BP%282%20born%20on%20same%20day%29%20%7D%3D1-%5B%5Cfrac%7B364%7D%7B365%7D%5D)

The probability that among ten students in a class, at least two of them share a birthday is 0.002.
Its depends on how many times you score during the game
in second game the number of points increases as a geometric sequence
with common ratio 2
so for example if you score ten times in first game you get 2000 points
if you score 10 times in Game 2 you score 2 * (2^10) - 1 = 2046 points
so playing 10 or more games is best with Game 2. Any less plays favours gane 1.